An inner product on a vector space $V$ ( over field $F$ ) is a function $V \times V \to F$, which associates to each pair of vectors $\bar{x},\bar{y}$ of $V$ a scalar quantity $\langle \bar{x},\bar{y}\rangle$, and satisfies the following properties : $$\langle \bar{x},\bar{x} \rangle>0$$
$$\langle \bar{x},\bar{y}\rangle=\langle \bar{y}, \bar{x} \rangle$$
$$\langle a\bar{x}+b\bar{y},\bar{z}\rangle=a\langle \bar{x},\bar{z}\rangle+b\langle \bar{y},\bar{z}\rangle.$$
Consider the subspace of all continuous functions on $[a,b]$ ; $\Bbb{C}$. Now, suppose we had some statement involving inner products that we wish to prove for vectors $\bar{x}, \bar{y} \in \Bbb{C}.$ For example, Let's say I choose to use the following inner product : $$\langle \bar{x},\bar{y}\rangle=\int_a^bx(t)y(t) dt.$$
If the statement is true for this inner product, is it then also true for all other inner products on $\Bbb{C}$?
Generally when working with inner products, there is a degree of choice as to what type of inner product is convenient for that particular situation - but what impact can this choice have ( if any )?
No, a statement true for one inner product is not necessarily true for another. For instance, you can use an innner product to define a length $$ \| x \| = \langle x,x \rangle^\frac12 $$
Under the standard inner product on $\Bbb R$ (defined by $(x, y) \mapsto xy$), we have $$ \| 2 \| = 2 $$ for instance. But under a different inner product, $(x, y) \mapsto 2xy)$, we have $$ \| 2 \| = 4. $$ Now that may seem pretty contrived, but it can be tough to sort out the contrived from the uncontrived.